document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=201660>Question 201660</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/9780470895023/algebrahomeworkh></A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> log(square root of 1-x) - log(square root of x+2)=logx<BR>\n" );
document.write( "First I sqared both sides and got log 1-x -log x+2=log x squared<BR>\n" );
document.write( "then i performed the quotient property of logs and got log(1-x/x+2)=log x squared then I got 1-x =x squared (x-2) then I got X cubed - 2x+ x-1=0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #151925 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=151925\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "In the first place, you didn't actually square both sides.  Squaring the log function does NOT result in the log of the square of the argument.  Furthermore, you are trying to say that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (a + b)^2 = a^2 + b^2\"> -- a statement that is false in general. Had you actually squared both sides of your original equation you would have a result that looked like:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \log^2(\sqrt{1 - x}) - \log(\sqrt{1 - x})\cdot\log(\sqrt{x + 2}) + \log^2(\sqrt{x + 2}) = \log^2(x)\">\r<BR>\n" );
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document.write( "I used to think that a mud fence was the gold standard for ugliness, but looking at that horror, I may change my mind.\r<BR>\n" );
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document.write( "Having said all of that, you actually stepped in a big pile of it and still came up smelling like a rose.\r<BR>\n" );
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document.write( "Remember your laws of logarithms.  First:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ \log_b(x^n) = n\log_b(x)\">\r<BR>\n" );
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document.write( "So back to the original equation:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ \log(\sqrt{1 - x}) - \log(\sqrt{x + 2}) = log(x)\">\r<BR>\n" );
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document.write( "Which, recalling that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large a^{\frac{1}{n}} = \sqrt[n]{a}\">, can be re-written:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\log(1 - x) - \frac{1}{2}\log(x + 2) = \log(x)\">\r<BR>\n" );
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document.write( "Multiplying both sides by 2:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ \log(1 - x) - \log(x + 2) = 2\log(x) = \log(x^2)\">\r<BR>\n" );
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document.write( "Which, coincidentally, and only coincidentally, is where you were after you say you squared both sides.  Like I said, you were lucky.\r<BR>\n" );
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document.write( "From there you correctly applied the rule that the difference of the logs is the log of the quotient:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ \log(\frac{1 - x}{x + 2}) = \log(x^2)\">\r<BR>\n" );
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document.write( "And then you correctly asserted that if two logs to the same base are equal, then their arguments must be equal:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ \frac{1 - x}{x + 2} = x^2\">\r<BR>\n" );
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document.write( "From here you had a small error.  Somehow in multiplying both sides by <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x + 2\">, you ended up with <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x - 2\"> on the right.  You should have arrived at:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ 1 - x = x^2(x + 2)\">\r<BR>\n" );
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document.write( "And arranging into standard form results in:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ x^3 + 2x^2 + x - 1 = 0\">\r<BR>\n" );
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document.write( "Applying the rational root theorem, the only possible rational roots are <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pm 1\">\r<BR>\n" );
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document.write( "I'll leave it for you to verify, but there are no rational roots.\r<BR>\n" );
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document.write( "That leaves us with applying the general solution of the cubic.\r<BR>\n" );
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document.write( "Fortunately, this cubic is monadic, so we can use a simplified form of the general solution:\r<BR>\n" );
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document.write( "For the cubic equation <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x^3 + ax^2 + bx + c = 0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x = u\ -\ \frac{p}{3u}\ -\ \frac{a}{3}\">\r<BR>\n" );
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document.write( "Where\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ p\ =\ b\ -\ \frac{a^2}{3}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ q\ =\ c\ +\ \frac{2a^3\ - \ 9ab}{27}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ \sqrt[3]{-\frac{q}{2}\ \pm\ \sqrt{\frac{q^2}{4}\ +\ \frac{p^3}{27}}}\">\r<BR>\n" );
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document.write( "The expression above for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large u\"> can generate up to three values (there are three cubic roots related by a factor which is one of the two complex cubic roots of one, and two square roots of any sign ; but these 6 expressions can generate only 3 pairs). This also applies to the final solutions for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">.\r<BR>\n" );
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document.write( "Lucky you -- you get to do your own arithmetic.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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